Q. 5.17 - A First Course in Probability

Question

The salaries of physicians in a certain specialty are approximately normally distributed. If $25$ percent of these

physicians earn less than $\backslash (180,000$ and $25$ percent earn more than $\backslash )320,000$, approximately what fraction earn

(a) less than $\backslash (200,000$?

(b) between $\backslash )280,000 \text{and} \$320,000$?

Step-by-Step Solution

Answer

(a) The fraction that earn less than $\$200,000$ is $0.3149$.

(b) The fraction that earn between $\$280,000$ and $\$320,000$ is 0.1348.

Part (a) Step 1. Given information.

Here, it is given that the salaries of physicians in a certain specialty are approximately normally distributed. 

$25\%$ of these physicians earn $< \$180,000$ and

$25\%$ of these physicians earn $> \$320,000$

Part (a) Step 2. Find the value of &#956; &#160; and &#160; &#963; .

Let $X$ be the random variable that represents the salary in thousand dollars.

So,

$$\begin{gathered} P(X<180) = 0.25 \\ P \left(\frac{X-\mu }{\sigma }<\frac{180-\mu }{\sigma }\right)=0.25 \\ P(Z<z) = 0.25 \end{gathered}$$

From standard normal table values, the critical value of $z$ for the one tail test and the corresponding cumulative area of $0.25$ is $-0.675$.

So,

$$\begin{gathered} z=-0.675 \\ \frac{180-\mu }{\sigma }=-0.675 \\ 180 = \mu -0.675\sigma ................... \left(1\right) \end{gathered}$$

and

$$\begin{gathered} P(X>320) = 0.25 \\ P \left(\frac{X-\mu }{\sigma }>\frac{320-\mu }{\sigma }\right)=0.25 \\ 1-P(Z\le z) = 0.25 \\ P(Z\le z) = 0.75 \end{gathered}$$

From standard normal table values, the critical value of $z$ for the one tail test and the corresponding cumulative area of $0.75$ is $0.675$.

So,

$$\begin{gathered} z=0.675 \\ \frac{320-\mu }{\sigma }=-0.675 \\ 320 = \mu +0.675\sigma ................... \left(2\right) \end{gathered}$$

From equation $\left(1\right) \text{and} \left(2\right)$, we get

$$\begin{gathered} 2\mu =500 \\ \therefore \mu =250 \end{gathered}$$

Substituting the value of $\mu$in equation $1$, we get

$$\begin{gathered} 180=250-0.675\sigma \\ 0.675\sigma = 70 \\ \therefore \sigma =103.704 \end{gathered}$$

Part (a) Step 3. Calculate the fraction of physicians that earn less than $ 200 , 000 .

$$\begin{gathered} P(X<200) = P \left(\frac{X-\mu }{\sigma }<\frac{200-\mu }{\sigma }\right) \\ = P \left(z<\frac{200-250}{103.704}\right) \\ = P \left(z<-0.48\right) \\ =0.3149 \end{gathered}$$

Therefore, the fraction of physicians that earn less than $\$200,000$ is $0.3149$.

Part (b) Step 1. Calculate the fraction of physicians that earn between $ 280 , 000 &#160; and &#160; $ 320 , 000 .

$$\begin{gathered} P(280<X<320) = P \left(\frac{280-\mu }{\sigma }<\frac{X-\mu }{\sigma }<\frac{320-\mu }{\sigma }\right) \\ = P \left(\frac{280-250}{103.704}<z<\frac{200-250}{103.704}\right) \\ = P \left(0.29<z<0.67\right) \\ = P\left(z<0.67\right) - P (z<0.29) \\ =0.7486 - 0.6138 \\ =0.1348 \\ =0.3149 \end{gathered}$$

Therefore, the fraction of physicians that earn between $\$280,000 \text{and} \$320,000$ is $0.3149$.